What is the last digit of \(1! + 2! + 3! + ... + 100!\) ?
Answer: D
Let's find the unit digits of the first few factorials.
\(1! = 1\) (unit digit is 1)
\(2! = 2\) (unit digit is 2)
\(3! = 6\) (unit digit is 6)
\(4! = 24\) (unit digit is 4)
\(5! = 120\) (unit digit is 0)
\(6! = 720\) (unit digit is 0)
For any n >= 5, n! will have 5 and 2 as factors, so its unit digit will be 0.
So we only need to sum the unit digits of the first four factorials.
Sum of unit digits = \(1 + 2 + 6 + 4 = 13\).
The unit digit of the entire sum is the unit digit of 13, which is 3.